Quadrilateral problem
 

A simple quadrilateral is inscribed in a semi-circle so that the diameter of the semi-circle is one side of the quadrilateral. (Simple just means that no two sides cross each other, i.e, no self-intersections.) If the other three sides of the quadrilateral have lengths 3, 4, and 5 units, what is the diameter/radius of the semi-circle?

[SPOILER]Using the fact that we have two pythagorean triangles (with the diameter as hypotenuse) and Ptolemy's equation (cyclic quadrilateral <=> AB x CD + AD x BC= AC x BD (where AC and BD are diagonals)) we stumble upon the following equation:

r^3-50r=120
[/SPOILER]

:newcat:

PS: forgot again the way to hide, seem to have not be posting for a while :blush:

I get the same equation, but with the diameter d rather than r.

One of the amusing things about this problem is the variety of ways to solve it. I have three solutions, all different from your interesting and elegant approach. However, I am still looking for a method of solution which elucidates the form of the answer.

For anyone who works out an algebraic solution, consider the generalization where 3, 4, and 5 are replaced by the variables a, b, and c. (I discovered the hard way that my algebra skills are actually better than my arithmetic skills!)

[QUOTE=philmoore;97344]I get the same equation, but with the diameter d rather than r.

One of the amusing things about this problem is the variety of ways to solve it. I have three solutions, all different from your interesting and elegant approach. However, I am still looking for a method of solution which elucidates the form of the answer.

For anyone who works out an algebraic solution, consider the generalization where 3, 4, and 5 are replaced by the variables a, b, and c. (I discovered the hard way that my algebra skills are actually better than my arithmetic skills!)[/QUOTE]

:smile: You are right philmoore and there are a variety of ways to solve the problem by cosidering the value of angles from the cosine formula and then proceeding or as Keyes Has mentioned by the Cyclic quad formula which she has rightly called Ptolemys Theorem.

The correct formula is d^3 -50d -120 = 0

where d is the diameter of the circle. But how about solving this cubic and giving us a near exact value ?

In all cubics one root is rational the other two roots are imaginary. So what is the rational value? Will it be plus or minus ?

I await with bated breath as its not as simple as it sounds.

Mally :coffee:

silly me :blush:
of course it should have been the equation for the diameter, but I decided to use the confusing variable r for diameter...

General equation should be

D^3-(a^2+b^2+c^2)D=2abc

where D (diameter) and a,b,c are the sides. Given the symmetry of the equation it does not matter in which order we label the sides.

PS: someone kindly hid my solution, but did not give me a clue as to how.
I vaguely remember having to put bars [] around something like hide and \hide
but the exact sequence escapes me

:cat:

[QUOTE=Kees;97364]PS: someone kindly hid my solution, but did not give me a clue as to how.
I vaguely remember having to put bars [] around something like hide and \hide
but the exact sequence escapes me

:cat:[/QUOTE]


To hide it use: [ SPOILER ] [ / SPOILER ] (without the spaces).

Dear Malcolm,

Kees (although it could be used for a girls name), refers to a boy here...
A rapid look at the equation reveals that 8<D<9

Approximately D = 8.05

Fiddling by moderator
 

[QUOTE=Kees;97364]silly me :blush:
of course it should h

PS: someone kindly hid my solution, but did not give me a clue as to how.
I vaguely remember having to put bars [] around something like hide and \hide
but the exact sequence escapes me

:cat:[/QUOTE]

:wink: Take a look at the bottom line of your post 2 and you will know who hid your Half baked solution. As is common in this thread once again it was itself half baked without the tip off like ATH's

Mally :coffee:

[QUOTE=philmoore;97283]A simple quadrilateral is inscribed in a semi-circle so that the diameter of the semi-circle is one side of the quadrilateral. (Simple just means that no two sides cross each other, i.e, no self-intersections.) If the other three sides of the quadrilateral have lengths 3, 4, and 5 units, what is the diameter/radius of the semi-circle?[/QUOTE]

diameter/radius = 2 for any circle ;)

Mistaken identity
 

[QUOTE=Kees;97367]Dear Malcolm,

Kees (although it could be used for a girls name), refers to a boy here...
A rapid look at the equation reveals that 8<D<9

Approximately D = 8.05[/QUOTE]

:smile:
Im sorry Kees as I deduced from your smiley and the avatar you have chosen both of which are decidedly feminine that you were of the opp. sex.
I am a full blooded male and a blush on either sex could be arousing to me though I am not that way inclined ! :wink:

Mally :coffee:

[QUOTE=mfgoode;97363]In all cubics one root is rational the other two roots are imaginary.[/QUOTE]This is not correct : it should be at least one root is a real number. For instance :
x[sup]3[/sup]-6*x[sup]2[/sup]+11*x-6 =0 has three real solutions, i.e; 1, 2 & 3
x[sup]3[/sup]-4*x[sup]2[/sup]+5*x-2=0 has two real solutions, i.e; 1 & 2 (2 being a double solution)
x[sup]3[/sup]-3*x[sup]2[/sup]+3*x-1=0 has one real solution, i.e; 1 (it is a triple solution)

Of course if the coefficients are randomly chosen reals, most of the time there will be only one real solution and two more can be found by using imaginary numbers.

Jacob